a,\(n_{Fe_2O_3}=\dfrac{1,6}{160}=0,01\left(mol\right)\)

\(m_{H_2SO_4}=19,6\%.200=39,2\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{39,2}{98}=0,4\left(mol\right)\)

\(PTHH:Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)

Mol:                0,01            0,03                0,01                  

Tỉ lệ:\(\dfrac{0,01}{1}< \dfrac{0,4}{3}\)⇒Fe₂O₃ pứ hết,H₂SO₄ dư

b,m_{dd sau pứ} = 200+1,6 = 201,6 (g)

\(C\%_{Fe_2\left(SO_4\right)_3}=\dfrac{0,01.400}{201,6}.100\%=1,98\%\)

\(C\%_{H_2SO_4dư}=\dfrac{\left(0,4-0,03\right).98}{201,6}.100\%=17,98\%\)

1.C

2.B

3.D

4.D

5.A

6.B

7.A

8.A

9.C

10.D

⇔68/17<x≤38/9

⇔4<x≤\(4\dfrac{2}{9}\)

như vậy mới phải chia ở qk hoàn thành

had lived-went-is living

1.taught

2.Do you like-will give

3.has lived-went-is living

4.watching-playing

5.doesn't know

6.rises-will disappear

7.playing-doing

8.going

9.will we do

10.Have your booked-will telephone

câu 3 sai

1.spend

2.things

3.fishing

4.home

5.join

6.when

7.singer

8.movies

9.watching

6.when mới đúng

a)4x²-9=0

⇔ (2x-3)(2x+3)=0

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

b)(x+5)²-(x-1)²=0

⇔ (x+5-x+1)(x+5+x-1)=0

⇔ 12(x+2)=0

⇔ x=-2

c)x²-6x-7=0

⇔ x²-7x+x-7=0

⇔ x(x-7)+(x-7)=0

⇔ (x-7)(x+1)=0

\(\Leftrightarrow\left\{{}\begin{matrix}x=7\\x=-1\end{matrix}\right.\)

d)(x+1)²-(2x-1)²=0

⇔ (x+1-2x+1)(x+1+2x-1)=0

⇔3x(2-x)=0

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)